2016 AMC 10A Problem 9

P1-10~2 minPrint

Sequences & Series Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$ ?

Show solution without answering

Key insight

The rows add up to N(N+1)/2 coins, so N times (N+1) is 4032. Since 63 squared is close to 4032, try 63 times 64, which is exactly 4032.

Solution

1. The rows hold $1, 2, 3, \ldots, N$ coins. The total $1 + 2 + \cdots + N$ equals $\dfrac{N(N+1)}{2}$ . (For example, $1 + 2 + 3 + 4 = 10 = \tfrac{4 \cdot 5}{2}$ .)
2. So
$\frac{N(N+1)}{2} = 2016, \qquad\text{which means}\qquad N(N+1) = 4032.$
3. $N$ and $N+1$ are next to each other, so their product is close to $N^2$ . That means $N$ is close to the square root of $4032$ .
4. Try nearby squares: $60^2 = 3600$ is too small, $64^2 = 4096$ is a bit too big. Test $63 \cdot 64 = 4032$ . It matches exactly, so $N = 63$ .
5. The question asks for the sum of the digits of $N$ : $6 + 3 = 9$ .

The answer is $\boxed{\textbf{(D)}\ 9}$ .

Why this works

"One more in each row" always adds up to a triangular number $\tfrac{N(N+1)}{2}$ . A product of two neighboring numbers sits just above $N^2$ , so a square-root estimate finds $N$ without solving a quadratic. Read the last line of the question carefully: it asks for a digit sum, not $N$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The total number of coins is $1 + 2 + \dots + N = \dfrac{N(N+1)}{2}$ , so
$N(N+1) = 2 \cdot 2016 = 4032.$

Since $N(N+1)$ is a product of consecutive integers, $N$ is just below $\sqrt{4032}$ . As $63^2 = 3969$ and $64^2 = 4096$ , try $N = 63$ : indeed $63 \cdot 64 = 4032$ .

The sum of the digits of $63$ is $6 + 3 = 9$ . The answer is $\boxed{\textbf{(D)}\ 9}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)